The equation of a circle $C$ is $x^2+y^2-8x+4y-16 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-8x) + (y^2+4y) = 16$ $(x^2-8x+16) + (y^2+4y+4) = 16 + 16 + 4$ $(x-4)^{2} + (y+2)^{2} = 36 = 6^2$ Thus, $(h, k) = (4, -2)$ and $r = 6$.